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15.12: Bases Ortonormales en Espacios Reales y Complejos

  • Page ID
    86577
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    Notación

    Transpose operador\(A^T\) voltea la matriz a través de su diagonal.

    \ [\ begin {array} {l}
    A=\ left (\ begin {array} {ll}
    a_ {1,1} & a_ {1,2}\
    a_ {2,1} & a_ {2,2}
    \ end {array}\ derecha)\\
    A^ {T} =\ left (\ begin {array} {ll}
    a_ {1,1} & a_ {2,1}\
    a_ {1,2} & a_ _ {2,2}
    \ end {array}\ right)
    \ end {array}\ nonumber\]

    Columna\(i\) de\(A\) es fila\(i\) de\(A^T\)

    Recall, producto interno

    \ [\ boldsymbol {x} =\ left (\ begin {array} {c}
    x_ {0}\\
    x_ {1}\\
    \ vdots\\
    x_ {n-1}
    \ end {array}\ derecha)\ nonumber\]

    \ [\ boldsymbol {y} =\ left (\ begin {array} {c}
    y_ {0}\\
    y_ {1}\\
    \ vdots\\
    y_ {n-1}
    \ end {array}\ derecha)\ nonumber\]

    \ [\ negridsymbol {x} ^ {T}\ negridsymbol {y} =\ left (\ begin {array} {cccc}
    x_ {0} & x_ {1} &\ ldots & x_ {n-1}
    \ end {array}\ derecha)\ left (\ begin {array} {c}
    y_ {0}\
    y_ {1}\\
    \ vdots\
    y_ _ {n-1}
    \ end {array}\ derecha) =\ suma_ { i}\ negridsymbol {x} _ {i}\ negridsymbol {y} _ _ {i} =\ langle\ negridsymbol {y},\ negridsymbol {x}\ rangle\ nonumber\]

    encendido\(\mathbb{R}^n\).

    Transposición hermitiana\(A^H\), transposición y conjugación

    \ [\ begin {array} {c}
    A^ {\ mathrm {H}} =\ overline {A^ {T}}\\ langle
    \ negridsymbol {y},\ negridsymbol {x}\ rangle=\ negridsymbol {x} ^ {\ mathrm {H}}\ negridsymbol {y} =\ sum_ {i}\ negridsymbol {x} _ i}\ bar {\ boldsymbol {y}} _ {i}
    \ end {array}\ nonumber\]

    encendido\(\mathbb{C}^n\).

    Ahora, dejemos\(\left\{b_{0}, b_{1}, \dots, b_{n-1}\right\}\) ser una base ortonormal para\(\mathbb{C}^n\)

    \ [\ begin {array} {c}
    i=\ {0,1,\ ldots, n-1\}\ izquierda\ langle b_ {i}, b_ {i}\ derecha\ rangle=1\
    \ izquierda (i\ neq j,\ izquierda\ langle b_ {i}, b_ {i}\ derecha\ rangle=b_ {j} ^ {H} b_ {i} =0\ derecha)
    \ end {array}\ nonumber\]

    Matriz de base:

    \ [B=\ left (\ begin {array} {cccc}
    \ vdots &\ vdots &\ vdots\\
    b_ {0} & b_ {1} &\ dots & b_ {n-1}\\\ vdots &
    \ vdots &\ vdots &\ vdots
    \ end {array}\ right)\ nonumber\]

    Ahora,

    \ [B^ {\ mathrm {H}} B=\ left (\ begin {array} {ccc}
    \ cdots & b_ {0} ^ {\ mathrm {H}} &\ cdots\
    \ cdots & b_ {1} ^ {\ mathrm {H}} &\ cdots\\
    &\ vdots\
    \ cdots & b_ {n-1} ^ {\ mathrm {H}} &\ ldots
    \ end {array}\ derecha)\ left (\ begin {array } {cccc}
    \ vdots &\ vdots &\ vdots\\
    b_ {0} & b_ {1} &\ cdots & b_ {n-1}\\
    \ vdots &\ vdots & &\ vdots
    \ end {array}\ derecha) =\ left (\ begin {array} {cccc}
    b_ {0} ^ {\ mathrm {H}} b_ {0}} & b_ {0} ^ {\ mathrm {H}} b_ {1} &\ ldots & b_ {0} ^ {\ mathrm {H}} b_ {n-1}\\
    b_ {1} ^ {\ mathrm {H}} b_ {0} & b_ {1} ^ {\ mathrm {H}} b_ {1} &\ ldots & b_ {1} ^ {\ mathrm {H}} b_ {n-1}\
    \ vdots & &\\
    b_ {n-1} ^ {\ mathrm {H}} b_ {0} & b_ {n-1} ^ {\ mathrm {H}} b_ {1} &\ ldots & b_ {n-1} ^ {\ mathrm {H}} b_ {n-1}
    \ end {array}\ derecha)\ nonumber\]

    Para base ortonormal con matriz de base\(B\)

    \[B^H=B^{-1} \nonumber \]

    (\(B^{T}=B^{-1} \text {in } \mathbb{R}^{n}\)in\(\mathbb{R}^n\))\(B^H\) es fácil de calcular mientras que\(B^{-1}\) es difícil de calcular.

    Entonces, para encontrar\(\left\{\alpha_{0}, \alpha_{1}, \ldots, \alpha_{n-1}\right\}\) tal que

    \[\boldsymbol{x}=\sum_{i} \alpha_{i} b_{i} \nonumber \]

    Calcular

    \[\left(\boldsymbol{\alpha}=B^{-1} \boldsymbol{x}\right) \Rightarrow\left(\boldsymbol{\alpha}=B^{H} \boldsymbol{x}\right) \nonumber \]

    Utilizando una base ortonormal nos libramos de la operación inversa.


    This page titled 15.12: Bases Ortonormales en Espacios Reales y Complejos is shared under a CC BY license and was authored, remixed, and/or curated by Richard Baraniuk et al..